Answer:
at = 0.18 m/s²
ac = 0.432 m/s²
a = 0.468 m/s² m/s²
Explanation:
Tangential acceleration is calculated as follows:
at = α*R Formula (1)
Centripetal or radial acceleration is calculated as follows:
ac =ω²*R Formula (2)
We apply the equations of circular motion uniformly accelerated :
ωf= ω₀ + α*t Formula (3)
Where:
α : angular acceleration (rad/s²)
t : time interval (s)
ω₀ : initial angular speed ( rad/s)
ωf : final angular speed ( rad/s)
R : radius of the circular path (m)
at: tangential acceleration, (m/s²)
ac: centripetal acceleration, (m/s²)
Data:
R = 0.3 m : radius of the flywheel
ω₀ = 0
α = 0.6 rad /s²
t = 2 s
Calculation of ωf at t= 2 s
We apply the Formula (3) :
ωf= ω₀ + α*t
ωf= 0 + ( 0.6)*(2)
ωf =1.2 rad/s
Calculation of the tangential acceleration
We apply the Formula (1)
at = α*R = ( 0.6)*(0.3) = 0.18 m/s²
Calculation of the radial acceleration
We apply the Formula (2)
ac =ω²*R = ( 1.2)²*(0.3) = 0.432 m/s²
Calculation of the resultant acceleration (a)
[tex]a= \sqrt{(a_{t})^{2}+(a_{c})^{2} }[/tex]
[tex]a= \sqrt{( 0.18)^{2}+(0.432)^{2} }[/tex]
a= 0.468 m/s²
The tangential acceleration of the flywheel is 0.18rad/s²
The radial acceleration of the flywheel is 0.432rad/s²
The resultant acceleration is 0.468rad/s²
The formula for calculating the tangential acceleration is expressed as:
[tex]a_t=\alpha R[/tex]
[tex]\alpha[/tex] is the angular acceleration
t is the time taken
Given the following parameters
[tex]\alpha[/tex] = 0.6rad/s²
R = 0.300secs
Substitute the given parameters into the formula:
[tex]a_t=0.6 \times 03\\a_t=0.18rad/s^2[/tex]
Hence the tangential acceleration of the flywheel is 0.18m/s²
Get the radial acceleration:
[tex]a_r=\omega^2r[/tex]
[tex]\omega[/tex] is the angular speed
Get the angular speed
[tex]\omega_f= \omega_0+\alpha t\\\omega_f= 0+0.6(2)\\\omega_f=1.2rad/s[/tex]
Get the radial acceleration:
[tex]a_r=\omega^2r\\a_r=1.2^2(0.3)\\a_r=0.432rad/s^2[/tex]
Hence the radial acceleration is 0.432rad/s²
Get the resultant acceleration:
[tex]a=\sqrt{a_t^2+a_r^2}\\a=\sqrt{0.18^2+0.432^2} \\a=\sqrt{0.0324+0.186624}\\a=\sqrt{0.219024} \\a=0.468rad/s^2[/tex]
Hence the resultant acceleration is 0.468rad/s²
Learn more here: https://brainly.com/question/3385759
